Question

A jet fighter at a height of $3000$ m from the ground, passes directly over another jet fighter at an instance when their angles of elevation from the same observation point are ${60}^{\circ }$ and ${45}^{\circ }$ respectively. Find the distance of the first jet fighter from the second jet at that instant. $(\sqrt{3}=1.732)$

Answer 1

Let $O$ be the point of observation.

Let $A$ and $B$ be the positions of the two jet fighters at the given instant when one is directly above the other.

Let $C$ be the point on the ground such that $AC=3000m$.

Given: $\angle AOC={60}^{\circ }$ and $\angle BOC={45}^{\circ }$

Let $h$ denote the distance between the jets at the instant.

Therefore, $BC=3000-h$.

In the right angled $△BOC$,

$\tan {45}^{\circ }=\frac{BC}{OC}$

$\Rightarrow OC=BC$ $(\because \tan {45}^{\circ }=1)$

Thus, $OC=3000-h$ .....(1)

In the right angled $△AOC,$

$\tan {60}^{\circ }=\frac{AC}{OC}$

$\Rightarrow OC=\frac{AC}{\tan {60}^{\circ }}=\frac{3000}{\sqrt{3}}$

$=\frac{3000}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=1000\sqrt{3}$ ...... (2)

From (1) and (2), we get,

$3000-h=1000\sqrt{3}$

$\Rightarrow h=3000-1000\times 1.732=1268$m

The distance of the first jet fighter from the second jet at that instant is $1268$ m.